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3.00 mol of PCl_5 kept in 1 L closed reaction vessel was allowed to attain equilibrium at 380 K. Calculate composition of the mixture at equilibrium. K_c=1.80 PCl_(5(g)) hArr PCl_(3(g)) + Cl_(2(g))

Answer» <html><body><p></p>Solution :`{:("Equilibrium reaction :",PCl_(5(g)) hArr , PCl_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>(g)) + , Cl_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>(g))),("Initial mol :", 3.0,0,0),("Change of mol :", -x,x,+x),("Mol at equilibrium :",(3-x),x,x),("mol" L^(-1) "at equilibrium :",(3-x)/1=(3-x),x/1=x, x/1=x):}` <br/> where , x=(<a href="https://interviewquestions.tuteehub.com/tag/mole-1100299" style="font-weight:bold;" target="_blank" title="Click to know more about MOLE">MOLE</a> of `PCl_5`) = (Mole of `PCl_3` and `Cl_2` ) <br/> So, `K_c=([PCl_3][Cl_2])/[[PCl_5]]` <br/> `therefore 1.80 = ((x)(x))/(3-x)=x^2/(3-x)` <br/> `therefore x^2`=1.80 (3-x)=5.4 -1.80x <br/> `therefore x^2+1.80x-5.4=0` <br/> `therefore` a=1 |b=1.80|c=-5.4<br/> `x=(-bpmsqrt(b^2-Delta4ac))/(<a href="https://interviewquestions.tuteehub.com/tag/2a-300249" style="font-weight:bold;" target="_blank" title="Click to know more about 2A">2A</a>)` <br/> `=(-1.80 pm sqrt((-1.8)^2-4(1)(-5.4)))/(2(1))` <br/> `=(-1.80 pm sqrt(3.24+21.6))/2` <br/>`=(-1.80 pm sqrt(24.84))/2=(-1.8 pm 4.984)/2`<br/> `therefore x=3.184/2` OR `(-6.084)/2` <br/>=1.592 M OR -3.042 M <br/> So, x=1.592and -3.42 is <a href="https://interviewquestions.tuteehub.com/tag/impossible-498943" style="font-weight:bold;" target="_blank" title="Click to know more about IMPOSSIBLE">IMPOSSIBLE</a> <br/> At equilibrium `(PCl_5)`=(3-x)=(3-1.592) =1.408 M<br/> `approx` 1.41 M<br/> equilibrium `[PCl_3]=[Cl_2]`= x= 1.592 M `approx` 1.59 M</body></html>


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